Mysql 如何删除重复记录并保留最近的记录？

我有一张这样的桌子：

Country_city('id', 'country_id', 'City_id')

并在其上复制我选择的记录：

SELECT 
    CONCAT(Country_id, '+', City_id), 
    COUNT(CONCAT(Country_id, '+', City_id)) 
FROM Country_City
GROUP BY CONCAT(Country_id, '+', City_id) 
HAVING COUNT(CONCAT(Country_id, '+', City_id)) > 1

结果如下：

CONCAT(Country_id, '+', City_id)  |  COUNT(CONCAT(Country_id, '+', City_id))
MA07+11                           |  48
MA07+13                           |  5
...

我试图删除它们我执行了以下操作：

DELETE FROM Country_City 
WHERE id NOT IN (
           SELECT MAX(id) 
           FROM Country_City 
           GROUP BY CONCAT(Country_id, '+', City_id)
           );

---

它不起作用；我的要求有什么问题吗？如何删除重复数据并保留一条记录？

请使用您的

CONCAT（国家/地区id，+'，城市id）

：

虽然此处可能不需要

CONCAT

，因此您也可以使用：

DELETE cc FROM Country_City cc
INNER JOIN
(
 SELECT Country_id,City_id,MAX(id) as id 
 FROM Country_City 
 GROUP BY Country_id,City_id
 HAVING COUNT(*) > 1           
) ccs on cc.Country_id = ccs.Country_id 
      and cc.City_id = ccs.City_id 
      and ccs.id != cc.id;

---

您是否尝试将别名放入

选择MAX（id）作为id

？。

DELETE cc FROM Country_City cc
INNER JOIN
(
 SELECT Country_id,City_id,MAX(id) as id 
 FROM Country_City 
 GROUP BY Country_id,City_id
 HAVING COUNT(*) > 1           
) ccs on cc.Country_id = ccs.Country_id 
      and cc.City_id = ccs.City_id 
      and ccs.id != cc.id;